Chemical F
ormulae o
f M
olecular C
ompounds Because molecular compounds are made from molecules, and molecules are made from atoms, molecular compounds are made from whole-­‐number ra:os of elements. Molecules of the same compound will always have the same ra:o of elements. For instance, water molecules always have two atoms of hydrogen and one atom of oxygen. The chemical formula of a molecular compound tells us the elements that make up one molecule. We write the number of atoms in each molecule as a subscript aCer each chemical symbol. H2O H2O2 CCl4 C2H5OH Chemical Formulae of Ionic Compounds The chemical formula of an ionic compound, also known as the unit formula, tells us the minimum number of ions needed to make that compound. An ionic compound will have a total charge of zero in its unit formula. For instance, sodium ions (Na+) will combine with the same number of chloride ions (Cl–) but calcium ions (Ca2+) will combine with twice as many Cl– ions. sodium chloride Na+Cl– calcium chloride Ca2+Cl2– calcium carbonate Ca2+CO32– Unlike molecular compounds, however, we almost never find ionic compounds in the form of their unit formulae. Instead, they are found as crystal laQces in solids or as separate ions in solu:ons. Wri:ng Chemical Equa:ons We describe a chemical reac:on by using a chemical equa6on. The number before the chemical formula tells us how many molecules (actually, how many moles, explained later) or unit formulae are needed. We must always have the same number of atoms (or moles of atoms) on both sides of the equa:on and the total of the charges must be the same on both sides. 2H2 + O2 2H2O 2P4 + 12Cl2 8PCl3 = P4 + 6Cl2 4PCl3 P4O10 + 6Ca2+(OH)–2 2Ca2+3(PO4)3–2 + 6H2O ( 4 x P; 6 x Ca; 22 x O; 12 x H) ( 4 x P; 6 x Ca; 22 x O; 12 x H) Atomic and Molecular Mass When we perform a chemical reac:on, we cannot (usually) count out individual atoms and molecules. Instead, we weigh out the quan:ty of the compound we need, just like cooking. In the laboratory we weigh out samples in quan::es from milligrams to grams. In industry, factories use quan::es from kilograms to tonnes. But how do we know how much to weigh out? Avogadro’s Number In 1811, an Italian chemist by the name of Avogadro proposed that equal volumes of gas at the same temperature contained equal numbers of par:cles. Using a fixed volume of gas, it was then possible to compare the weights of the same numbers of molecules of different elements and then calculate the rela:ve weights of their atoms. Ini:ally, hydrogen was given the weight of ‘1’ because it was the lightest element. A number called the Avogadro number was fixed as the number of atoms of hydrogen in 1 gram of hydrogen gas. To make more accurate measurements, however, the Avogadro number has since been changed to the number of atoms in 12 grams of carbon-­‐12. Measurements show that this number is about 6.022 x 1023 or 602,200,000,000,000,000,000,000. hydrogen gas (H2) No. of par:cles = 6.022 x 1023 weight = 2.02 g oxygen (O2) No. of par:cles = 6.022 x 1023 weight = 32.0 g chlorine (Cl2) No. of par:cles = 6.022 x 1023 weight = 70.0 g Argon (Ar) No. of par:cles= 6.022 x 1023 weight = 40.0 g The Mole and Rela:ve Atomic Masses When Avogadro’s number is used for coun:ng things, we call it a mole (unit: mol). So, one mole of helium gas would contain 6.022 x 1023 atoms. The mole is an essen:al concept in chemistry because it is the standard number used for coun:ng atoms and molecules. This is because it is more convenient to count in moles rather than count the atoms or molecules themselves, i.e. it is easier to say 2 moles of carbon than 12.044 x 1023 atoms of carbon. This picture shows one mole of seven kinds of elements. In the top row, it shows 79.9 g of bromine (Br), 27.0 g of aluminium (Al), 200.6 g of mercury (Hg), 63.5 g of copper (Cu) and in the bohom row, it shows 32.1 g of sulfur (S), 65.4 g of zinc (Zn) and 55.8 g of iron (Fe). All of these samples contains the same number of atoms, i.e. roughly 6.022 x 1023 atoms. The mass of one typical mole of atoms of an element is known as its rela6ve atomic mass (Ar). Periodic tables usually list the rela:ve atomic masses of the elements together with their chemical symbols and atomic numbers. Rela:ve atomic masses have units of grams per mole, (wrihen as g /mol or gmol–1) Atomic Mass Units and Rela:ve Atomic Mass Electrons are 2000 :mes less massive than nucleons (protons or neutrons) so, in chemistry, we ignore the mass of electrons when calcula:ng the mass of an atom. 1H, 1 nucleon, 1 atomic mass unit 12C, 12 nucleons, 12 atomic mass units Since one mole is defined as the number of atoms in a pure sample of 12 g of 12C (12 nucleons), one mole of nucleons weighs exactly 1 g. But tables of rela:ve atomic mass list the rela:ve atomic mass (Ar) of hydrogen as 1.0079 gmol–1, carbon as 12.0107 gmol–1 and chlorine as 35.453 gmol–1. How can we have frac:onal atomic masses if we cannot have frac:ons of a nucleon? This is because rela:ve atomic masses are based on the weight of a typically available mole of an element, which will contain mixtures of different isotopes. The rela:ve atomic mass of carbon is 12.0107 gmol–1 because a typical sample of 6.022 x 1023 atoms of carbon contains mostly 12C with a :ny amount of 13C and 14C. The atomic mass of chlorine is 35.453 gmol–1 because a typical mole of chlorine contains a mixture of about two thirds 35Cl and about one third 37Cl. Calcula:ng Moles from Ar Values Using Avogadro’s number and the rela:ve atomic mass of an element, we can easily calculate the number of atoms in a sample (or moles of atoms in a sample) from its weight, and vice versa. No. of moles of atoms in a sample of = element mass of the sample rela:ve atomic mass of the element It is important to remember that one mole of hydrogen gas contains 6.022 x 1023 hydrogen molecules but, since hydrogen molecules (H2) are made of two hydrogen atoms each, one mole of hydrogen molecules, weighing 2.0 g, would contain two moles of hydrogen atoms. Likewise, 6.0 g of carbon is 0.5 mol of carbon atoms, 56 g of iron contains 1 mol of iron atoms, and 14 g of nitrogen gas (N2) contains 1 mol of nitrogen atoms but only 0.5 mol of nitrogen molecules. 32g of solid sulphur (S8) contains 1 mol of sulphur atoms but only 0.125 mol of S8 molecules (one thousandth of a mole is called a millimole (mmol) so 0.125 mol is also wrihen as 125 mmol). Rela:ve Molecular Masses As we have just seen, we can easily calculate the mass of one mole of oxygen molecules (O2), because it must be twice the mass of one mole of oxygen atoms, i.e. 16.0 g x 2 = 32.0 g. We can do the same for any combina:on of atoms and hence calculate the rela6ve molecular mass (Mr) for any compound. E.g. one molecule of water (H2O) contains two atoms of hydrogen and one atom of oxygen, therefore the mass of one mole of water molecules will equal the the masses of two moles of hydrogen atoms plus the mass of one mole of oxygen atoms, i.e. 2.0 g + 16.0 g = 18.0 g. Mr O2 = (2x16) gmol–1 = 32 gmol–1 (Ar O = 16 gmol–1) Mr N2 = (2x14) gmol–1 = 28 gmol–1 (Ar N = 14 gmol–1) Mr H2O = (2x1) gmol–1 + 16 gmol–1 = 18 gmol–1 (Ar H = 1 gmol–1) Mr CaCO3 = 40 gmol–1 + 12 gmol–1 + (3x16 gmol–1) = 100 gmol–1 (Ar Ca = 40 gmol–1) (Note: most of you will have used rela:ve atomic masses accurate to zero decimal places for calcula:ons in high school. From now on, you will have to use an appropriate number of significant figures for your calcula:ons. When calcula:ng an answer for a test answer (or real experiments) always determine the appropriate number of significant figures for your calcula:ons) Calcula:ng Moles — Examples Now we can see that we can also apply the equa:on for calcula:ng the number of moles of an element to calcula:ng the number of moles of a compound. No. of moles in sample = mass of sample ______________ Mr of sample This equa:on means that we can equate the number of moles (and hence molecules) in a sample to its weight. We can then measure out the correct weight of each compound or element when performing chemical reac:ons or determine the number of molecules that come out of reac:ons. 1.0 g 16 g ________ ________ = 0.056 mol = 56 mmol = 0.50 mol 1.0 g of H2O => 16 g of O2 => –1
–1
18 gmol 32 gmol 25 g = 0.25 mol = 250 mmol 25 g of CaCO3 => ________ 100 gmol–1 Of course, we can also perform this calcula:on in reverse. If we know how many moles of a compound we want, we can calculate how much to weigh out: 2 mol of NaOH => 2 mol x 40.0 gmol–1 = 80.0 g 0.25 mol of CuSO4 => 0.25 mol x 160.0 gmol–1 = 40.0 g Using Moles — Example Let’s suppose that we want to make 10 g of calcium iodide, an ionic solid with the unit formula of CaI2. First, we calculate the Mr of CaI: Mr of CaI2 = 40.0 gmol–1 + (2x127) gmol–1 = 294 gmol–1 Next, we calculate how many moles cons:tutes 10 g of calcium iodide: 10 g ________ = 0.034 mol ( = 34 mmol) –1
294 gmol From the unit formula, CaI2, we can see that every mole of CaI2 contains one mole of Ca and two moles of I. Therefore, to make 0.034 mol (10g) of CaI2 we need to start with one equivalent of Ca and two equivalents of I, i.e. 0.034 mol of Ca, and 0.068 mol of I: Ca: 0.034 mol x 40 gmol–1 = 1.4 g I: 0.068 mol x 127 gmol–1 = 8.6 g We can now see that we need to react 1.4 g of Ca with 8.6 g of I to make 10 g of CaI2. These calcula:ons are very important when calcula:ng how much of different elements and compounds we need when performing chemical reac:ons. Solu:ons and Concentra:on Usually in chemistry, it is necessary to use compounds dissolved in liquids. When we dissolve compounds in liquids, the dissolved compound is called a solute, the liquid is called a solvent and both together are a solu6on. When we use solu:ons, for instance hydrogen chloride (HCl) in water, we need to calculate the concentra6on. In cooking we use weights and volumes, e.g. 1g of salt in 1L of water, but in chemistry it is more convenient to use moles and volume. The molar mass of HCl is 36.5 gmol–1, so if we dissolve 36.5 g of HCl in enough water to make a solu:on of 1L (or 1 dm3), the concentra:on of HCl is 1 molar (1 M or 1 mol dm–3). 18.25g of HCl in 1 dm3 of solu:on would be 0.5 M and 365g of HCl in 1 dm3 of solu:on would be 10 M. 10 M is a very concentrated solu:on and the highest concentra:on that can be achieved with HCl in water. Water is a very common solvent and water solu:ons are known as aqueous solu:ons. Using Concentra:ons We want to add 0.1 moles of HCl to a reac:on using a 1M solu:on. How much (in terms of volume) of the solu:on do we add to our reac:on? Concentra:on = moles volume Volume = moles concentra:on Volume = 0.1 mol 1 mol dm–3   From this calcula:on we can see that we need 0.1 dm3 = 100 ml.   We use square brackets to mean ‘the concentra:on of…’. For instance [HCl] = 5 M means ‘the concentra:on of HCl is 5 molar’. Using Moles is an Essen:al Skill Make sure you thoroughly understand the concept of moles and how to use them. Using moles is an essen:al skill in any chemical calcula:on that uses quan::es. In other words, any :me you are solving a problem related to ‘how much’ in chemistry, you will have to use moles and rela:ve molecular or rela:ve atomic masses. If you do not understand how to perform calcula:ons that use moles to calculate masses and concentra:ons (and vice versa) be sure to ask. DO NOT WAIT UNTIL EXAM TIME. Past Exam Problems A)  Calculate how much (grams) NaOH can be obtained starting with 10 g of Na. Write the required mass of O2 (assuming all of the oxygen originally came from O2).
B) i) What mass of sodium hydrogen carbonate (NaHCO3) is required to make 150 ml of a 5.0M solution? iii) What is the minimum amount of 5.0M NaHCO3 solution we need to completely react with 18 ml of 3.0 M HNO3 solution? (Assume 1 mole of NaHCO3 reacts with 1 mole of HNO3). Answers A)  Calculate how much (grams) NaOH can be obtained starting with 10 g of Na. Write the required mass of O2 (assuming all of the oxygen originally came from O2).
(First, convert 10 g Na into moles of Na atoms) Ar Na = 23 g (2 s.f.) => ________ 10 g Na = 0.435 mol Na 23 gmol–1 (1 mole of NaOH contains 1 mole of Na so maximum possible moles of NaOH is also 0.435 mol) Mr NaOH = 23 gmol–1 + 16 gmol–1 + 1.0 gmol–1 = 40 gmol–1 => 40 gmol–1 × 0.435 mol = 17.4 g = 17 g NaOH (2 s.f.) Note: you can also calculate this method by no:ng that the Mr of NaOH is 1.7 :mes the Ar of Na, so you could mul:ply 10g of Na by 1.7 and get 17 g NaOH that way. This method works but most students (and some :red professors) make simple arithme:c mistakes when taking this shortcut. I always recommend doing the calcula:on the long way round, i.e. going through the number of moles of reactants and products. Answers B) i) What mass of sodium hydrogen carbonate (NaHCO3) is required to make 150 ml of a 5.0M solution? (First, calculate how many moles is contained in 150 ml of 5M solu:on) 0.150 dm3 × 5 mol dm–3 = 0.750 mol (Now convert the moles into mass) Mr NaHCO3 = 23 gmol–1 + 1.0 gmol–1 + 12 gmol–1 + (3 × 16 gmol–1) = 84 gmol–1 => mass of required NaHCO3 is 0.750 mol × 84 gmol–1 = 63.0 g = 63 g (2 s.f.) iii) What is the minimum amount of 5.0 M NaHCO3 solution we need to completely react with 18 ml of 3.0 M HNO3 solution? (Assume 1 mole of NaHCO3 reacts with 1 mole of HNO3). (The long way via moles) 0.018 dm3 × 3 mol dm–3 = 0.054 mol => 0.054 mol = 0.0108 dm3 = 0.011 dm3 (2 s.f.) = 11 ml (either dm3 or ml are acceptable) ________ 5 mol dm–3 (The quick way is very easy since 1 mole of NaHCO3 reacts with 1 mole of HNO3) 0.018 dm3 × 3 mol dm–3 = x dm3 × 5 mol dm–3 => x dm3 = 18 ml × _________ 3 mol dm–3 = 10.8 ml = 11 ml (2 s.f.) 5 mol dm–3 Reading and Problems Reading Chapter 2: 2-­‐4, 2-­‐5, 2-­‐6 Chapter 3: 3-­‐1, 3-­‐6 (molarity) Problems Chapter 2: 28, 32, 38, 77 Chapter 3: 8, 62